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  • Confused! kindly explain, - Electrostatics - JEE Main-10

If the potential function is given by  V=4x+3y, then the magnitude of electric field intensity at the point (2, 1) will be

  • Option 1)

    11

  • Option 2)

    5

  • Option 3)

    7

  • Option 4)

    1

 

Answers (1)

best_answer

As we learned

 

In space -

E_{x}=\frac{-dv}{dx}  ,  E_{y}=\frac{-dv}{dy}    ,  E_{z}=\frac{-dv}{dz}

-

 

 By using i.e.,  E=\sqrt{E_{x}^{2}+E_{y}^{2}};E_{x}=-\frac{dV}{dx}=-\frac{d}{dx}(4x+3y)=-4

E_{y}=-\frac{dV}{dy}=-\frac{d}{dy}(4x+3y)=-3

\therefore E=\sqrt{(-4)^{2}+(-3)^{2}} = 5N/C

 


Option 1)

11

Option 2)

5

Option 3)

7

Option 4)

1

Posted by

Avinash

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